chapter 3 5. derive a relation between drift velocity and

SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut 1

CHAPTER 3

CURRENT ELECTRICITY

1. Define electric current. Give its SI

unit.

Ans: Current is the rate of flow

of electric charge.

I = dt

dq[ or I = q/t ]

SI unit is ampere (A), 1A=1C/s

2. Define current density ( J). Give its

SI unit.

Ans: It is the current flowing

through unit area of cross section

of a conductor.

J = A

I

S.I unit is A/m2

3. Define drift velocity (vd).

Ans: In the absence of an external

electric field, inside a conductor the

free electrons are in random motion;

and so their average velocity is zero.

When an electric field is applied

the electrons move opposite to the

electric field and get an average

velocity.

The average velocity acquired by

an electron in the presence of an

external electric field is called

drift velocity.

4. Define relaxation time.

Ans: Relaxation time is the

average time interval between

two successive collisions.

5. Derive a relation between drift

velocity and current.

Ans: Consider a conductor of length

‘’ and number density of electrons ‘n’.

[n = number of electrons per volume]

The distance travelled by electron in dt

time = vddt

The number of free electrons in the

distance vddt

= Avddt × n

The total charge crossing the area A in

dt time

dq= neAvddt

Current I = dt

dq

I = dneAv dt

dt= neAvd

J = I / A

J = dneAv

A

6. Derive a relation between drift

velocity (vd ) and relaxation time (τ).

Ans: Let vi be the initial velocity of an

electron just after a collision. The

velocity of the electron after the

relaxation time ‘τ’ is given by,

vd = vi + aτ

=i

eEv

m


But just after a collision electron can

be taken at rest ie, vi = 0

vd = eE

m

dv

= e E

m

7. Define mobility

Ans: Mobile charge carriers are

responsible for conductivity.

In metals, electrons are the charge

carriers.

In electrolytes both +ve and –ve ions

are charge carriers.

In semiconductors, conduction is

partially by electrons and partially by

holes.

Mobility ‘µ’ is defined as the ratio

of magnitude of the drift velocity

to electric field strength.

µ = | |

dv eE e

E E m

SI unit of mobility is CmN-1s-1 or

m2V-1S-1.

8. Distinguish between emf and

potential difference.

Ans:

1. Emf is the difference in

potential between the

terminals of a cell, when no

current is drawn from it.

Potential difference is the

difference in potential between

the terminals of a cell or between

any two points in a circuit when

current is drawn from the cell.

2. Emf exists only between the

terminals of the cell. But

potential difference exists

throughout the circuit.

3. Emf is the cause and

potential difference is the

after effect.

4. Emf is always greater than

potential difference.

9. State Ohm’s law.

Ans: Ohm’s law states that ‘at

constant temperature the current

flowing through a conductor is

directly proportional to potential

difference between the ends of

the conductor’.

Current α Potential difference.

I α V I = 1

VR

R Resistance of the conductor.

V = IR

10. What is meant by resistance?

Ans: It is the ability of a

conductor to oppose electric

current.

SI unit of resistance is ohm.

11. What are the factors which affect

the resistance of a conductor?

Ans:

1. Nature of the material

of the conductor.


2. Length of the conductor

‘’ (directly

proportional)

3. Area of cross-section ‘A’

(Inversely proportional)

4. Temperature (directly

proportional)

12. Write the equation for the

resistance of a conductor in terms of

its length and area of cross section.

Ans: R ∝

R ∝1

𝐴

R ∝𝑙

𝐴

R=𝝆𝒍

𝑨

𝜌 The resistivity of the material of

the conductor.

S.I. unit of Resistivity ohm-metre (Ω -

m)

13. Define resistivity of the material

of a conductor.

Ans: Resistivity,

𝝆=𝑹𝑨

𝒍

Put A = 1 m2, = 1m

𝜌 = 𝑅.1

1 =R

“Resistivity of the material of a

conductor is defined as the resistance

of the conductor having unit length

and unit area of cross section.

14. Define conductance (C)

Ans: Conductance C = 1

𝑅

Conductance is the reciprocal of

resistance.

SI unit = 1

Ω = ohm-1 = mho.

15. Define conductivity (𝝈)

Ans:

Conductivity 𝜎 = 1

𝜌

Conductivity is the reciprocal of

resistivity.

SI unit of conductivity is Ω-1m-1or mho

m-1

16. What are the factors which

affect resistivity?

Ans: i. Nature of the material of

the conductor.

ii. Temperature

17. Derive the relation for resistivity

in terms of relaxation time.

Ans:

d

d

2

2

We know, I neAv

eEand v

m

eE ne AEI neA

m m

VBut we have E

ne AVI

m

2

2

2

V m

I ne A

mR

ne A

m

A ne A

2

2

Resistivity,

m

ne

Conductivity,

ne

m


18. A wire of resistance 4𝛀 is drawn (a) to twice its original length

(b) to thrice its original length. Calculate the new resistance in each case 19[P]. A wire of resistance 4R is bent in to the form of a circle. What is the effective resistance between the ends of diameter?

20[P]. A copper wire is in the form of a cylinder and has a resistance R. It is stretched till its thickness reduces by half of its initial size. Find its new resistance in terms of R.

21[P].The voltage current graphs for two resistors of the same material and same radii with lengths L1 and L2 are shown in the figure. If L1>L2, state with reason, which of these graphs represents voltage current change for L1.

22[Q]. Two wires of equal lengths, one of copper and the other of manganin,

have the same resistance. Which wire is thicker? 23. Why copper is used as for

making connecting wires?

Ans: Copper has low resistivity.

24. Why Nichrome is used as

heating element of electrical

devices?

Ans: Nichrome has

i) High resistivity

ii) High melting point.

25. Why aluminium wires are

preferred for overhead power

cables?

Ans: Aluminium has low resistivity. It

is cheaper and lighter.

26. What are Ohmic and Non-

ohmic substances.

Ans:

Ohmic substances

They are the substances which obey

ohms law. For these substances V-I

graph is linear.

Eg: Metals

Non – ohmic substances

They are the substances which do not

obey ohm’s law. For these substances

V-I graph is nonlinear.


Eg:-Semiconductors, diodes,

vacuum tubes, electrolytes etc.

27. Ohm’s law is not a universal law.

Explain.

Ans: All materials do not obey Ohm’s

law. Metals obey ohms law while

semiconductors, electrolytes, diodes

etc. do not obey Ohm’s law. So Ohm’s

law is not a universal law.

28. Explain how resistance depends

on temperature.

Ans: When temperature increases,

resistances of materials change.

R2 = R1[1 + α (T2-T1)]

R1 resistance at the reference

temperature T1.

R2 resistance at temperature T2.

∝ Temperature coefficient of

resistance.

Unit of ∝ is 0C-1 or K-1

The value of ∝ is positive for metals

negative for semiconductors and nearly

zero for insulators.

29. Why materials like constantan

and manganin are used to make

standard resistances?

Ans: The temperature coefficient of

resistance of these materials is nearly

zero. Therefore, their resistances do

not change considerably with

temperature. Moreover these materials

have high resistivity.

30[P]. At room temperature (27.00C) the resistance of a heating element is 100 𝜴. What is the temperature of the element if the resistance is found to be 117 𝜴. Given that the temperature coefficient of the material of the resistor is 1.70 x 10-4 0C-1 ?

31[P]. A silver wire has a resistance of 2.1 𝜴 at 27.50C and a resistance of 2.7𝜴 at 1000C. Determine the temp coefficient of resistivity of silver.

32. Give the codes for different

coloured rings of carbon resistors.

Ans:

Colour coding of carbon

resistors

Colour Code Tolerance

Black 0

Brown 1

Red 2

Orange 3

Yellow 4

Green 5

Blue 6

Violet 7

Grey 8

White 9

Gold -1 5%

Silver -2 10%

No

colour

20%


33. Determine the resistance of the

carbon resistor in the following

figure.

Ans:-

Resistance R = (47×102±5%)Ω

34[P]. Determine the resistances of

the following carbon resistors.

35. Determine the resistance of a

resistor with the following colours

for the rings: Blue, Grey, Black, No

colour.

Ans:-R = (68 × 100 ± 20%)Ω

= (68 ± 20%) Ω

36. What is the use of colour coding.

Ans: Colour coding of a carbon

resistance is used to find the value of

resistance.

37. Derive expressions for the

effective resistance when three

resistors are connected in (i) series

(ii) parallel

Ans:

(i) series

Consider resistances R1, R2 and R3

connected in series with a voltage V.

In series circuit the current is the

same, but the voltages across

different resistors are different.

The applied voltage,

V = V1 + V2 + V3 ---- (1)

But V = IRs

V1 = IR1, V2 = IR2 and V3 =

IR3

(1) IRs = IR1 + IR2 + IR3

IRs = I (R1+R2+R3)

Rs = R1 + R2+ R3

If there are ‘n’ resistors.

Rs = R1 + R2 + R3 + ……… + Rn

ii) parallel

Consider three resistances R1, R2

and R3 connected in parallel.

In parallel circuit voltage across

each resistor is the same but the

currents are different.

I = I1 + I2 + I3 …………. (2)

I = p

V

R


I1 = 1

V

R I2 =

2

V

R I3 =

3

V

R

(2) p

V

R =

1

V

R+

2

V

R+

3

V

R

If there are n resistors

1 2 3

1 1 1 1

pR R R R+ …………….. +

1

nR

38[P]. (a) Three resistors 1 𝛺, 2𝛺 𝑎𝑛𝑑 3𝛺 are combined in series. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 12V and negligible internal resistance, obtain the potential drop across each resistor.

39[P]. (a)Three resistors 2 𝛺, 4𝛺 𝑎𝑛𝑑 5𝛺 are combined in parallel. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 20V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery.

40[P]. Given n resistors each of resistance ‘R’. How will you combine them to get the (i) maximum (ii)

minimum effective resistance? What is the ratio of the maximum to minimum effective resistance?

41[P]. Determine the equivalent resistance of the network shown in figure.

42[P]. Find the equivalent resistance between A and B.

43. Define internal resistance of a

cell.

Ans: In the external circuit, the current

flows from the positive terminal of the

cell to the negative terminal. But inside

the cell the current flows from –ve

terminal to the +ve terminal.

“Internal resistance of a cell is the

resistance offered by the electrolyte

and electrodes of the cell”.


Effective resistance = R + r

Current I = E

R r

I(R + r) = E IR + Ir = E

But IR = V V Terminal Voltage

V + Ir = E

Ir = E – V

E V

rI

44. What are the factors which affect

the internal resistance of a cell?

Ans: (i) The nature of the

electrolyte and the electrodes.

(ii) The distance between

the electrodes.

(iii) Temperature

45. It is easier to start a car engine on

a warm day than a chilly day. Why?

Ans: Internal resistance of a car battery

decreases on a warm day due to

increase in temperature. Therefore,

more current flows to the spark plug.

46[P]. The storage battery of a car has an emf of 12V. If the internal resistance of a battery is 0.4𝜴, what is

the maximum current that can be drawn from the battery?

47[P]. A battery of emf 10V and internal resistance 3𝜴 is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resistor? What is the terminal voltage of the battery?

48. Derive expressions for the

current due to a combination of

cells.

Ans:

i) Series

Consider n cells connected in series

with a resistor R.

Total emf= E +E + E + ……. + E = nE

Total Internal resistance

= r + r + r + …………… + r = nr

Effective resistance = R + nr

Current I = tan

emf nE

resis ce R nr

ii) Parallel

Consider n cells connected in parallel

with an external resistance.


Total internal resistance

49. State Kirchhoff’s rules.

Ans:

1st rule (Current rule or Junction

rule)

Kirchhoff’s first rule states

that “In a closed circuit the

algebraic sum of the currents

meeting at a junction is zero”

In other words,

The total current entering a junction

is equal to the total current leaving

the junction.

I1 + I2 + I3 – I4 – I5 = 0

Or I1 + I2 + I3 = I4 + I5

Kirchhoff’s Ist rule is a statement of law of

conservation of charge.

2nd rule [Voltage Rule or Loop rule

or mesh rule]

Kirchhoff’s second rule states

that ‘In a closed circuit the

algebraic sum of the voltages is

zero’.

In the closed circuit ABFGA

I2R2 + I1R1= E

Similarly for the mesh [closed circuit]

BCDFB,

I3R3 -I2R2 = 0

Kirchhoff’s second rule is a statement of law

of conservation of energy.

Wheatstone’s Bridge

50. What is the use of a

Wheatstone’s bridge?

Ans: It is used to find the resistance

of a given wire.


51. Derive the balancing condition

of Wheatstone’s bridge. Explain

how the unknown resistance can be

determined.

Ans:

Circuit Details

Consider 4 resistances P, Q, R

and S connected in the form of a

bridge. A cell of emf ‘E’ is connected

between the terminals A and C. A

galvanometer is connected between the

terminals B and D.

Balancing Condition

Applying Kirchhoff’s 2nd rule in the

loop ABDA,

I1P +IgG - I2R = 0 …. (1)

Again in the mesh BCDB,

I3Q - I4S - IgG = 0 …………. (2)

The resistance ‘S’ is adjusted so that

current through the galvanometer is

made zero. ie.,Ig = 0

Then, I1 = I3 and I2 = I4

(1) I1P - I2R = 0;

I1P = I2R ………… (3)

(2) I3Q - I4S = 0;

I3Q = I4S ……….. (4)

(3)/(4) 1 2

3 4

I P I R

I Q I S

But when the bridge is balanced, I1=I3

and I2= I4

P R

Q S

This is the balancing condition of

Wheatstone’s bridge.

Procedure

The wire whose resistance is to

be determined (Q) is connected across

B and C. The value of resistance S is

adjusted so that the current through the

galvanometer is zero. Now the bridge

is balanced and we have P R

Q S

Since the values of P, Q, R and S are

known, we can calculate the value of Q

using the equation PS

QR

52[P]. Calculate the equivalent resistance between X and Y of the following network.

53[P]. Determine the current in each branch of the network shown in figure

.


Meter Bridge

54. What is the use of a meter

bridge?

Ans: It is used to find the resistance of

a given wire.

55. Using a neat circuit diagram

explain the circuit details, principle

of a meter bridge. And give the

procedure to find an unknown

resistance.

Ans:

Circuit Diagram

Circuit Details

Meter bridge consist of a one

meter long resistance wire (made of

constantan or manganin) fixed on a

wooden board. A cell of emf E and a

key are connected between the

terminals A and B. A jockey is

connected to the terminal ‘C’ through

a galvanometer. The unknown

resistance is connected in the left gap

and a resistance box is connected in the

right gap.

Principle

It works on the Wheatstone’s bridge

principle. At the balancing condition

P R

Q S

,

no current flows through the

galvanometer.

X r

R (100 )r

X

R 100

X = Rl

(100 l)

RX

100

Procedure

The key is closed. A suitable

resistance (say 1Ω ) is taken from the

resistance box. The jockey is moved

from A towards B, until the

galvanometer shows the zero

deflection. The balancing length (AJ =

) is measured. Unknown resistance

X is determined using the equation

X = R

100.

X is determined for different

values of R (say 2Ω, 3Ω, -----). Now

the experiment is repeated after

interchanging X and R.

Resistivity

Resistivity RA

L ,

Here, R = X, A = πr2

2X r

L

L is the length and r is

the radius of the given wire whose

resistance is to be determined.

56. (a) In a meter bridge the balance point is found to be at 39.5cm from


the end A, when the resistor Y is of 12.5𝛺. Determine the resistance of X. Why are the connections between resistors in a meter or Wheatstone bridge made of thick copper strips?

(b) Determine the balance point of the bridge if X and Y are interchanged.

(c) What happens if the galvanometer and cell are interchanged at the balancing point of the bridge? Would the galvanometer show any current?

57. What are the uses of a

potentiometer?

Ans:

Potentiometer is used

(i) To compare the emf of

two cells

(ii) To find the internal

resistance of a cell

58. Explain the principle of a

potentiometer.

Ans:

Consider a resistance wire of uniform

area of cross section carrying a current

I

The potential difference across

length of the wire is given by,

V = IR

V = I r , r is the resistance per

unit length of the wire

V = (Ir) V = k

V α , where k=Ir

Principle When a steady current (I) flows

through a wire of uniform area of

cross section, the potential

difference between any two

points of the wire is directly

proportional to the length of the

wire between the two points.

59. Using a neat connection

diagram, explain how potentiometer

can be used to compare the emf of

two cells.

Ans:

Circuit details

The primary circuit consists of

potentiometer, cell of emf ε, key K and

a rheostat.

The secondary circuit consists of two

cells ε1 and ε2, two-way key,

galvanometer, high resistance and

Jockey.

Connection diagram

Procedure


The key K in the primary circuit is

closed. The cell є1 is brought into the

circuit by putting the key. Jockey is

moved from A towards B until the

galvanometer shows zero deflection.

The balancing length AJ= is found

out. Now by the principle of

potentiometer,

ε1 1 ….. (1)

The cell ε1 is replaced by the cell ε2

by putting the key. Again balancing

length 2 is found out.

Therefore, ε2 2 …….. (2)

(1) ÷ (2)

1 1

2 2

If one of

the cells is a standard cell ( say ε1),

we can determine the emf of the

other

.

22 1

1

60. Using a neat connection

diagram explain how potentiometer

can be used to find the internal

resistance of a cell.

Ans:

circuit details

The primary circuit consists of a

potentiometer, cell of emf є′, key K′

and a rheostat. The secondary circuit

consist of a cell of emf є, a

galvanometer, Jockey, a resistance box

R and key K.

Connection Diagram

Procedure

The key ‘K′’ in primary circuit is

closed. The balancing length 1 is

found out.

ε 1 ………… (1)

Now the key K in the secondary is

closed. The balancing length 2 is

found out. Then we can write,

V 2

2

R

R r

……….. (2)

(1) ÷ (2) 1

2R

R r

1

2

R r

R

1

2

1

2

1 2

2

RR r

Rr R

R Rr

r = 1 2

2

R( )

Using this equation we can

calculate the internal resistance of the

given cell.


61. Why potentiometer is preferred

over voltmeter for measuring emf of

a cell?

Ans: If we measure emf of a cell using

a voltmeter, a small current is drawn by

the voltmeter. So we will not get the

actual emf of the cell. But if we

measure the emf using a potentiometer,

while measuring emf, since null

deflection method is employed, no

current flows from the cell. So we will

get the actual emf of the cell.

62[P]. In a potentiometer arrangement, a cell of emf 12.5V gives a balance point at 35.0cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0cm, what is the emf of the 2nd cell?

63[P]. In the potentiometer circuit shown, the balance point is at X.

State with reason, where the balance point will be shifted, when (a) Resistance R is increased,

keeping all parameters unchanged.

(b) Resistance S is increased, keeping R constant.

(c) Cell P is replaced by another cell whose emf is lower than that of cell Q.

64[P]. A potentiometer wire of length 100cm has a resistance of 10 ohms.

It is connected in series with a resistance R and a cell of emf 2V and of negligible internal resistance. The circuit is as shown below:

(a) What is the resistance of 40cm

length of the potentiometer wire?

(b) If a source of emf 10 millivolts is balanced by length of 40 cm of the potentiometer wire, find the value of the external resistance R.

(c) While performing an experiment on potentiometer what precautions (at least two) one must observe in the laboratory?


65. State Joule’s law of heating.

Ans: Joule’s Law states that “the

heat produced in a current

carrying conductor is

proportional to the square of

the intensity of electric

current, the resistance of the

conductor and the time for

which the current flows”.

H = I2Rt

S.I. unit of heat is joule (J)

66. Define electric power.

Ans: Electric power is defined as the

electric energy consumed per second

SI unit of electric power is watt (W)

67P. A light bulb is rated at 125W for a 250 V a.c supply. Calculate the resistance of the bulb. 68P. Two electric bulbs one 25W and other 100W, which one has greater resistance?

69. Which is the commercial unit of

electric energy? What is its relation

with joule.

Ans: The commercial unit of electric

energy is kilo watt hour.

1kWh=1000W x 3600s =3.6 x 106J

.

chapter 3 5. derive a relation between drift velocity and

Documents